K3 surfaces with 9 cusps in characteristic p

TL;DR

This paper investigates K3 surfaces with nine cusps over fields of characteristic p not equal to 3, establishing their relation to abelian surfaces and exploring their position in the supersingular locus.

Contribution

It proves that such K3 surfaces admit a triple cover by an abelian surface and characterizes which abelian surfaces produce these K3 surfaces.

Findings

Each K3 surface with 9 cusps admits a triple cover by an abelian surface.

Characterization of abelian surfaces with automorphisms leading to K3 surfaces.

Analysis of the occurrence of these K3 surfaces in the supersingular locus.

Abstract

We study K3 surfaces with 9 cusps, i.e. 9 disjoint $A_2$ configurations of smooth rational curves, over algebraically closed fields of characteristic $p\neq 3$. Much like in the complex situation studied by Barth, we prove that each such surface admits a triple covering by an abelian surface. Conversely, we determine which abelian surfaces with order three automorphisms give rise to K3 surfaces. We also investigate how K3 surfaces with 9 cusps hit the supersingular locus.

Tables

Type	$e$	$d$	$\eta$	char $p>0$
I	$e_0$	$1$	$1$	$e\mid 2$
II	$e_0$	$2$	$\frac{3}{4}$	$2e\mid 2$
III	$e_0$	$2$	$\frac{1}{4}$	$e\mid 2$
IV	$2e_0$	$d$	$\frac{1}{2}$	$e_{0}d\mid 2$

Equations

$$L:=(A_2^9{)}^{\prime }\subset \mathrm{NS}(X).$$

$$\Lambda ={\Lambda }_{\text{K3}}=U^3+E_8^2$$

$$\Lambda ={\Lambda }_{p,\sigma }=\mathrm{NS}(X)$$

$$A_{\Lambda }={\mathbb{F}}_p^{2\sigma }.$$

$$A_L\cong {\mathbb{F}}_3^r.$$

$$H\cong A_L,$$

$$q_L=-q_{L^{\perp }}{|}_H.$$

$$q_L=-q_M$$

$$y^2+y=x^3,$$

$$x\mapsto \omega x,y\mapsto y$$

$$\left(\begin{array}[]{cc}0&\iota\\ id&\iota\end{array}\right)$$

$$L\hookrightarrow \Lambda ={\Lambda }_{p,2}$$

$$N=L^{\perp }\left(\frac{1}{p}\right),$$

$$N\cong U(3)+A_2.$$

$${\mathbb{F}}_3^3\cong H\subset A_{L^{\perp }}\cong {\mathbb{F}}_p^4\times {\mathbb{F}}_3^3$$

$$q_{L^{\perp }}{|}_H=p\cdot q_N=\{\begin{array}{ll}{q}_N& \text{ if p≡1 mod }3,\\ -q_N& \text{ if p≡−1 mod }3,\end{array}$$

$$q_L=-q_{L^{\perp }}{|}_H$$

$$v=\frac{1}{3}\sum _{i=1}^9(C_i+2C_i^{\prime })$$

$$\begin{array}[]{ccccc}\tilde{A}&\to&\hat{A}&\to&A\\ \downarrow&&&&\downarrow\\ \tilde{X}&\to&X&\to&A/\langle\sigma\rangle\end{array}$$

$$0⟶E⟶A\stackrel{f}{⟶}A/E⟶0$$

$$\dim {\mathrm{H}}^0(A/⟨\sigma ⟩,{\Omega }_{A/⟨\sigma ⟩}^1)\ge 1,$$

$$\begin{array}[]{|c||c|c|c|c|c|c|}\hline\cr\text{Type}&e&e_{0}&d&\eta&\dim_{{\bf Q}}D&\rho(A)\\ \hline\cr\text{(I-i)}&1&1&1&1&1&1\\ \hline\cr\text{(I-ii)}&2&2&1&1&2&2\\ \hline\cr\text{(II)}&1&1&2&\frac{3}{4}&4&3\\ \hline\cr\text{(III-i)}&1&1&2&\frac{1}{4}&4&1\\ \hline\cr\text{(III-ii)}&2&2&2&\frac{1}{4}&8&2\\ \hline\cr\text{(IV-i)}&2&1&1&\frac{1}{2}&2&1\\ \hline\cr\text{(IV-ii)}&2&1&2&\frac{1}{2}&8&4\\ \hline\cr\text{(IV-iii)}&4&2&1&\frac{1}{2}&4&2\\ \hline\cr\end{array}$$

$$\mathrm{End}(A){\otimes }_{\mathbf{Z}}{\mathbf{Z}}_p⟶{\mathrm{End}}_{{\mathbf{Z}}_p}(T_p(A))$$

Full text

Partially supported by JSPS Grant-in-Aid for Scientific Research (B) No. 15H03614

1. Introduction

In two papers from the 1990’s [Barth1], [Barth2], Barth studied complex K3 surfaces with 9 cusps, i.e. with 9 disjoint $A_{2}$ configurations of smooth rational curves. Barth’s arguments were of topological nature, using a triple cover by some suitable abelian surface. In this paper, we follow a more algebraic approach which lends itself to investigate the same problem over any algebraically closed field $k$ of characteristic $p\neq 3$ (which we fix throughout this paper). This enables us to detect several interesting phenomena; in particular, we also include the Zariski K3 surfaces in characteristics $p\equiv-1$ mod $3$ from [KS]. Combined with explicit calculations for abelian surfaces (in positive characteristic) and the characteristic-free divisibility results for certain divisor classes from [S-nodal], we prove the following results:

Theorem 1.1**.**

If $X$ is a K3 surface with 9 cusps, then $X$ admits a triple covering by an abelian surface with an automorphism of order $3$.

Theorem 1.2**.**

If $X$ is a supersingular K3 surface with 9 cusps, then

• •

either $X$ is the supersingular K3 surface of Artin invariant $\sigma=1$,

• •

or $X$ has Artin invariant $\sigma=2$ and $p\equiv-1\;{\rm mod}\;3$.

Both theorems are supported by ample examples, starting from suitable abelian surfaces with an automorphism of order $3$. In fact, for an abelian surface to admit a triple K3 quotient is quite restrictive, both in the simple and non-simple case:

Proposition 1.3**.**

Let $A$ be an abelian surface such that A is isogenous to $E_{1}\times E_{2}$ with elliptic curves $E_{i}$ ($i=1,2$). Assume $E_{1}$ is not isogenous to $E_{2}$. Let $\sigma$ be an automorphism of $A$ of order 3. Then, $A/\langle\sigma\rangle$ is not birationally equivalent to a K3 surface.

In comparison, simple abelian surfaces are quite delicate to treat as we shall explore in Sections 6, LABEL:s:simple. In the context of this paper, it turns out that ordinarity enters as an essential ingredient:

Theorem 1.4**.**

Let $A$ be a simple ordinary abelian surface with an automorphism $\sigma$ of order 3. Assume that $\sigma$ is not a translation. Then, the quotient surface $A/\langle\sigma\rangle$ is birationally equivalent to a K3 surface.

Simple abelian surfaces in positive characteristic turn out to be quite hard to exhibit explicitly, especially in characteristic $2$. Therefore we conclude the paper with an explicit one-dimensional family of K3 surfaces with 9 cusps valid in any characteristic $\neq 3,5$ such that the generic covering abelian surface is simple (and we also provide an alternative family covering characteristic $5$).

Remark 1.5*.*

Many of our arguments also work in characteristic zero, but to ease the presentation we decided to restrict to the positive characteristic case.

2. Lattice theory for K3 surfaces with 9 cusps

Let $X$ be an algebraic K3 surface over an algebraically closed field $k$ of characteristic $p\neq 3$. Assume that $X$ contains 9 disjoint $A_{2}$ configurations of smooth rational curves. Then we have to determine the primitive closure of the resulting sublattice in ${\rm{NS}}(X)$:

[TABLE]

From general lattice theory (see e.g. [Nikulin]), we know that $L$ is determined by some isotropic subgroup $H$ of the discriminant group $G=(A_{2}^{\vee})^{9}/A_{2}^{9}$. Here the latter space is identified with the vector space $\mathbb{F}_{3}^{9}$, so the given problem can be analysed using coding theory. In [Barth2], Barth achieves this by showing (topologically over $\mathbb{C}$) the following lemma:

Lemma 2.1**.**

Any non-zero vector in $H$ has length 6 or 9.

Proof.

The same holds true in arbitrary characteristic since isotropic vectors of length $3$ would yield a vector in $L\setminus A_{2}^{9}$ of square $-2$, a contradiction to [S-nodal] (which is valid in any characteristic). ∎

In order to determine $L$, it will be instrumental to work out a suitable reference lattice $\Lambda$ into which $L$ embeds primitively. If $X$ has finite height, then it is known by work of Deligne that $X$ lifts to characteristic zero with a full set of generators of ${\rm{NS}}(X)$. Hence we can take $\Lambda$ to be the standard even unimodular lattice of rank $22$ and signature $(3,19)$,

[TABLE]

(to which $H^{2}(Y,\mathbb{Z})$ of any complex K3 surface $Y$ is isomorphic). On the other hand, if $X$ is supersingular, say of Artin invariant $\sigma$, then we may just take

[TABLE]

the unique even hyperbolic lattice of rank $22$ and discriminant group

[TABLE]

What unifies both variants is that they have the same rank while being prime to $3$ in the sense that, by assumption, the discriminant is not divisible by $3$. In comparison, $L$ is also prime to $p$ since it has discriminant $-3^{r}$, where $r=9-2\cdot|H|$. More precisely, $L$ has discriminant group

[TABLE]

By construction, $L$ embeds primitively into $\Lambda$. Since $L$ and $\Lambda$ are relatively prime in the above terminology, the orthogonal complement $L^{\perp}$ admits a subgroup $H\subseteq A_{L^{\perp}}$ such that not only the discriminant groups are isomorphic,

[TABLE]

but also the discriminant forms agree up to sign:

[TABLE]

In particular, $H$ and $A_{L}$ share the same length (i.e. minimum number of generators). Presently this is $r$ by (2.1), and on the other hand, the length is a priori bounded by the rank of $L^{\perp}$, i.e. $r\leq 4$.

Lemma 2.2**.**

$L$* is an overlattice of $A_{2}^{9}$ of index $27$, determined uniquely up to isometries by its discriminant form*

[TABLE]

for $M=U(3)+A_{2}(-1)$.

Proof.

The construction of $L$ follows exactly the lines of [Barth2], just using the existence of some isomorphism (2.2) and Lemma 2.1. In particular, this shows that $L$ is unique up to isometries. In loc. cit. it was also proved that $L^{\perp}$ inside $\Lambda_{\text{K3}}$ is isometric to $M=U(3)+A_{2}(-1)$. Since the shape of $L$ does not depend on the characteristic, the statement on the discriminant forms is always valid. ∎

Remark 2.3*.*

The above argument also shows as in [Barth2] that the subgroup $H$ of $G$ contains a vector of length $9$. This will be quite useful in the proof of Theorem 1.1.

3. Proof of Theorem 1.2

We are now ready to prove Theorem 1.2. To support it, we recall the following two well-known constructions of K3 surfaces with 9 cusps (cf. Katsura [K], for instance).

Example 3.1**.**

Let $E$ be an elliptic curve defined by

[TABLE]

and let $\sigma$ be an automorphism of $E$ given by

[TABLE]

with $\omega$ a primitive cube root of unity. Then, $\sigma\times\sigma^{2}$ is an automorphism of the abelian surface $A=E\times E$ of order 3 and the quotient surface $A/\langle\sigma\times\sigma^{2}\rangle$ is birationally equivalent to a K3 surface with 9 cusps. Note that in case $p\equiv 1~{}({\rm mod}~{}3)$, $A$ is ordinary, and in case $p\equiv-1~{}({\rm mod}~{}3)$, $A$ is supersingular (since the same holds for the elliptic curve $E$).

Example 3.2**.**

Let $E$ be an elliptic curve, and we set $A=E\times E$. Let $\sigma$ be the automorphism of $A$ defined by

[TABLE]

where $\iota$ is the inversion of $E$. Then $\sigma$ has order 3 and the quotient surface $A/\langle\sigma\rangle$ is birationally equivalent to a K3 surface with 9 cusps. Note that in case $E$ is ordinary, $A$ is also ordinary, and in case $E$ is supersingular, $A$ is also supersingular.

Accidentally, we treated the case of supersingular K3 surfaces of Artin invariant $\sigma=2$ in characteristic $p\equiv-1$ mod $3$ in [KS]. Namely we proved that all these K3 surfaces are Zariski (i.e. unirational, admitting an inseparable covering by $\mathbb{P}^{2}$ of degree $p$) by exploiting exactly the structures imposed by a configuration of 9 disjoint $A_{2}$’s.

In order to prove Theorem 1.2, it remains to treat the cases of Artin invariants $\sigma>2$ as well as $\sigma=2$ in characteristic $p\equiv 1$ mod $3$.

3.1. Artin invariant $\sigma>2$

In the previous section, we bounded the length of $A_{L}$ by considering the orthogonal complement $L^{\perp}$ inside the reference lattice $\Lambda$ (which was coprime to $L$). Here we can argue along similar lines for $\Lambda$ itself (cf. [KS, Thm. 6.1]). Namely, for the same reason as above, the discriminant group $A_{\Lambda}\cong\mathbb{F}_{p}^{\,2\sigma}$ has to be supported on $A_{L^{\perp}}$. But again, $L^{\perp}$ has rank $4$, so $\sigma\leq 2$ as claimed.

3.2. Artin invariant $\sigma=2$ in characteristic $p\equiv 1$ mod $3$

For reasons to become clear in a moment, we omit the restriction on the characteristic for the time being. That is, we just assume that $\sigma=2$, and for simplicity that $p>3$ (because for computations with even lattices it is often easier to exclude $p=2$). Suppose that $L$ admits a primitive embedding

[TABLE]

and let $L^{\perp}$ denote the orthogonal complement as before. We have seen above that $A_{\Lambda}$ is supported on $A_{L^{\perp}}$. Presently, this means that $A_{L^{\perp}}$ has $p$-length $4$, i.e.  $L^{\perp}$ is $p$-divisible as an even lattice (since $p>2$). We can thus scale $L^{\perp}$ by $1/p$ and obtain an even hyperbolic lattice

[TABLE]

of rank $4$ and discriminant $-27$ (the same as the discriminant of $L$ up to sign). We claim that

[TABLE]

To see the claim, we impose a duality in the spirit of [Kondo-Shimada] to derive the even hyperbolic lattice $N^{\vee}(3)$ of same rank $4$, but discriminant $-3$. The invariants are small enough to infer that $N^{\vee}(3)\cong U+A_{2}$. Now the claim follows by applying the duality again (since $A_{2}^{\vee}(3)\cong A_{2}$).

To conclude, we return to the subgroup

[TABLE]

from (2.2). The discriminant form on $H$ can be read off from (3.1) as follows:

[TABLE]

(the quadratic forms taking values in $\mathbb{Q}/2\mathbb{Z}$). Recall that gluing $L$ to $L^{\perp}$ along $H$ requires exactly that

[TABLE]

Independent of $p$, we already know that $q_{L}=q_{N}$ since $N=M(-1)$, see the proof of Lemma 2.2. Note that this asserts the case $p\equiv-1$ mod $3$ in (3.2). Hence the other alternative, with $p\equiv 1$ mod $3$, can only persist (for some $p$) if $q_{N}=-q_{N}$. But this is absurd – for instance, it would imply that $N$ glues to itself to give an even unimodular lattice, but this would have signature $(2,6)$, contradiction. This completes the proof of Theorem 1.2. ∎

4. Proof of Theorem 1.1

With these lattice theoretic preparations, it is not hard to give a proof of Theorem 1.1. Starting from a K3 surface $X$ containing 9 disjoint $A_{2}$ configurations of smooth rational curves, we not only have the sublattice $L$ inside ${\rm{NS}}(X)$ from Lemma 2.2, but we are also equipped with a vector $v$ of length $9$ inside the discriminant group $G$ of $A_{2}^{9}$ which in fact is integral, i.e. belongs to $L$ (see Remark 2.3). Explicitly, $v$ may be represented as

[TABLE]

where the $C_{i},C_{i}^{\prime}$ are the smooth rational curves supporting the nine $A_{2}$ configurations (up to exchanging the two curves). Following classical theory (e.g. [Miranda]), this divisor determines a triple covering of $X$ which we can use to our advantage. Indeed, we can proceed exactly as in [KS, §5], so we just give the rough outline of the construction:

1. (1)

blow up the intersection points $C_{i}\cap C_{i}^{\prime}$ to get $\tilde{X}$; 2. (2)

switch to the smooth triple covering $\tilde{A}$; 3. (3)

minimalize to $A$ by first blowing down the strict transforms of the $C_{i},C_{i}^{\prime}$ and then those of the exceptional curves in $\tilde{X}$; 4. (4)

check using the classification of algebraic surfaces that $A$ is an abelian surface.

For each step, the arguments from [KS] go through – regardless of the characteristic and of the question whether $X$ is supersingular or not. Automatically, the triple covering endows $A$ with an automorphism $\sigma$ of order $3$ such that $X$ can be recovered as minimal desingularization of the quotient $A/\langle\sigma\rangle$, and this completes the proof. (The following diagram of maps is only reproduced for the convenience of the reader.) ∎

[TABLE]

5. Non-simple abelian surfaces with automorphism of order 3

This section provides a proof of Proposition 1.3, so we let $A$ be an abelian surface such that A is isogenous to $E_{1}\times E_{2}$ with elliptic curves $E_{i}$ ($i=1,2$). We assume that $E_{1}$ is not isogenous to $E_{2}$ and that $A$ admits an automorphism $\sigma$ of order 3.

If $\sigma$ is fixed-point free, then it is clear that the quotient surface is either abelian of hyperelliptic. It remains to consider the case where $\sigma$ has a fixed point on $A$, say $P$. By our assumption, we have an exact sequence

[TABLE]

with an elliptic curve $E$. Let $F$ be a fiber of $f$ such that $F$ passes through the fixed point $P$. Then we have $\sigma(F)\cap F\ni P$. If $\sigma(F)\neq F$, then $\sigma(F)$ would be a multi-section of $f$ and we would have an isogeny from $\sigma(F)$ to $A/E$. Therefore, $A$ would be isogenous to $F\times A/E$ with $F$ isgenous to $A/E$, a contradiction to our assumption. It follows that $F=\sigma(F)$. We put $f(F)=Q$. Then, we have $3F=f^{*}(3Q)$ and, by Riemann–Roch, $3Q$ is a very ample divisor on $A/E$. Therefore, the linear system $\mid 3F\mid$ gives the morphism $f$. Since $F$ is invariant under the action of $\sigma$, $\sigma$ acts on the vector space $L(3F)$. Therefore, $\sigma$ induces an action on $A/E$ fixing $Q$.

Suppose first that $\sigma$ does not act as the identity on $F$ nor on $A/E$. Thus both elliptic curves admit an automorphism of order $3$, but this curve is unique up to isomorphism (j-invariant zero), so $F\cong A/E$, contradiction.

Suppose that $\sigma$ acts as identity on $A/E$. Then we have a morphism $A/\langle\sigma\rangle\longrightarrow A/E$. Therefore, we have the dimension $q(A/\langle\sigma\rangle)\geq 1$ of the Albanese variety of $A/\langle\sigma\rangle$. In particular, $A/\langle\sigma\rangle$ cannot be birationally equivalent to a K3 surface.

Suppose that $\sigma$ acts as identity on $F$. Since $F$ is non-singular, there exist local coordinates $x$, $y$ such that $\sigma(x)=x$ and $\sigma(y)=\omega y$ with $\omega$, a cube root of unity (it may be 1). Applying the same argument to any fixed point of $\sigma$, we see that the quotient surface $A/\langle\sigma\rangle$ is non-singular. Therefore, we have ${\rm H}^{0}(A/\langle\sigma\rangle,\Omega_{A/\langle\sigma\rangle}^{1})\cong{\rm H}^{0}(A,\Omega_{A}^{1})^{\langle\sigma\rangle}$. Since we have a natural isomorphism $m_{P}/m_{P}^{2}\cong{\rm H}^{0}(A,\Omega_{A}^{1})$ and $\dim(m_{P}/m_{P}^{2})^{\langle\sigma\rangle}\geq 1$, we obtain

[TABLE]

and again $A/\langle\sigma\rangle$ cannot be a K3 surface. This concludes the proof of Proposition 1.3. ∎

Corollary 5.1**.**

Let $A$ be an abelian surface with $p$-rank 1. Then, there exists no automorphism $\sigma$ of order 3 on $A$ such that $A/\langle\sigma\rangle$ is birationally equivalent to a K3 surface.

Proof.

If $A$ is simple, we will show this corollary in the next section. If $A$ is non-simple, $A$ is isogenous to a product $E_{1}\times E_{2}$ of two elliptic curves $E_{i}$ ($i=1,2$). Since the $p$-rank of $A$ is 1, one of $E_{i}$’s is ordinary and the other is supersingular. Therefore, $E_{1}$ is not isogenous to $E_{2}$. Hence, the result follows from Propositon 1.3. ∎

6. Endomorphism algebras of simple abelian surfaces

As before, we let $k$ be an algebraically closed field of characteristic $p>0$. We summerize here some results by Mumford on the endomorphism algebras of simple abelian surfaces ([M] Section 21, Theorem 2) over $k$. Let $A$ be a simple abelian surface and let ${\rm End}(E)$ be the endomorphism ring of $A$. We denote by $A_{n}$ the (reduced) $n$-torsion group of $A$. We set $D={\rm End}^{0}(E)={\rm End}(E)\otimes_{{\bf Z}}{\bf Q}$. Then, $D$ is a central simple division algebra. We denote by $K$ the center of $D$ and by $K_{0}$ the subfield of $K$ which is fixed by the Rosati involution. We put $[D:K]=d^{2}$, $[K:{\bf Q}]=e$ and $[K_{0}:{\bf Q}]=e_{0}$. We also put $S=\{x\in D\mid x^{\prime}=x\}$. It is known that $\dim_{{\bf Q}}S$ is equal to the Picard number $\rho(A)$ of $A$. We put $\eta=\frac{\dim_{{\bf Q}}S}{\dim_{{\bf Q}}D}$. Then, Mumford gave the following table for the possible numerical invariants of $D$.

Using this list, we get the following detailed list.

[TABLE]

We will show that the cases (III-ii), (IV-ii) and (IV-iii) cannot occur for a simple abelian surface $A$. We denote the $p$-adic Tate module of $A$ by $T_{p}(A)$. First we show the following lemma (see also [M], Section 19, Theorem 3).

Lemma 6.1**.**

Let $A$ be a simple abelian surface. Then, the natural homomorphism

[TABLE]

is injective.

Proof.

In dimension 2, $A$ is supersingular if and only if the $p$-rank of $A$ is 0. Since $A$ is simple, $A$ is not supersingular (cf. Oort [Oo]). Therefore, the rank of $T_{p}(A)$ is either 1 or 2. Since $A$ is simple, the kernel of a non-zero endomorphism $f$ is a finite group scheme. Therefore, for a large positive integer m, the induced homomorphism $f:A_{p^{m}}\longrightarrow A_{p^{m}}$ is not the zero-map. Therefore, the natural homomorphism ${\rm End}(A)\otimes_{{\bf Z}}{\bf Z}_{p}\longrightarrow{\rm End}_{{\bf Z}_{p}}(T_{p}(A))$ is injective. ∎

Lemma 6.2**.**

In the list above, the cases (III-ii), (IV-ii) and (IV-iii) cannot occur.

Proof.

In cases (III-ii) and (IV-ii), we have $\dim_{{\bf Q}_{p}}D\otimes_{{\bf Q}}{\bf Q}_{p}=8$. On the other hand, $\dim_{{\bf Q}_{p}}{\rm End}_{{\bf Q}_{p}}(T_{p}(A)\otimes_{{\bf Z}_{p}}{\bf Q}_{p})$ is equal to 1 or 4, according to the $p$-rank of $A$ = 1 or 2, which is impossible by Lemma 6.1. In case (IV-iii), since we have $\dim_{{\bf Q}}D=4$, the $p$-rank of $A$ should be 2 and $\dim_{{\bf Q}_{p}}{\rm End}_{{\bf Q}_{p}}(T_{p}(A)\otimes_{{\bf Z}_{p}}{\bf Q}_{p})=4$, and we have $D\otimes_{\bf Q}{{\bf Q}_{p}}\cong{\rm End}_{{\bf Q}_{p}}(T_{p}(A))\otimes_{{\bf Z}_{p}}{\bf Q}_{p}$. However, $D\otimes_{\bf Q}{{\bf Q}_{p}}$ is commutative and ${\rm End}_{{\bf Q}_{p}}(T_{p}(A))\otimes_{{\bf Z}_{p}}{\bf Q}_{p}$ is non-commutative, a contradiction. ∎

By the list above, we have the following corollary.

Corollary 6.3**.**

For simple abelian surfaces, we have $\rho(A)\leq 3$.

Note how this fits together with the classic result of Shioda–Mitani [SM] that a complex abelian surface $A$ with $\rho(A)=4$ is isomorphic to a product of elliptic curves.